Which of the following order is not correct?

Options

A $\mathrm{MeBr} > \mathrm{Me}_{2} \mathrm{CHBr} > \mathrm{Me}_{3} \mathrm{CBr} > \mathrm{Et}_{3} \mathrm{CBr}$ \left(\mathrm{S}_{\mathrm{N}}2\right)
B $\mathrm{Me}_{3} \mathrm{CBr} > \mathrm{Me}_{2} \mathrm{CHBr} > \mathrm{Me}_{2} \mathrm{CH.CH}_{2} \mathrm{Br} > \mathrm{MeCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}.$ \left(\mathrm{E}_{2}\right)
C $\mathrm{PhCH}_{2} \mathrm{Br} > \mathrm{Ph} \mathrm{CHBrMe} > \mathrm{PhCBrMe}_{2} > \operatorname{PhCBrMePh}$ \left(\mathrm{S}_{\mathrm{N}}1\right)
D $\mathrm{MeI} > \mathrm{MeBr} > \mathrm{MeCl} > \mathrm{MeF} \left(\mathrm{S}_{\mathrm{N}}2\right)$

Correct answer

C. $\mathrm{PhCH}_{2} \mathrm{Br} > \mathrm{Ph} \mathrm{CHBrMe} > \mathrm{PhCBrMe}_{2} > \operatorname{PhCBrMePh}$ \left(\mathrm{S}_{\mathrm{N}}1\right)

Step-by-step solution

The more is the stability of intermediate carbonium ion, the more is the chance of

\mathrm{S}_{\mathrm{N}} 1

mechanism. The intermediates obtained will be

\mathrm{Ph}^{+} \mathrm{C} \mathrm{H}_{2}(\mathrm{i}), \mathrm{Ph} \mathrm{C} \mathrm{H}-\mathrm{Me}(\mathrm{ii})\mathrm{Ph} \mathrm{C}-\mathrm{Me}_{2}\left(\right.

iii)

, \mathrm{Ph} \mathrm{C}^{+} \mathrm{MePh}

(iv) The stabilty is of the order iv

> \mathrm{iii} > \mathrm{ii} > \mathrm{i}

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